Given an integer N (N > 2), the task is to generate a binary string of size N that consists of equal numbers of “10” & “01” subsequences and also the string should contain at least one ‘0’ and one ‘1’
Note: If multiple such strings exist, print any.
Examples:
Input: 4
Output: 0110
Explanation : Here, 0110 string consists equal count of ’01’ and ’10’
subsequence and it consists at least one occurrence of 0 as well as 1.Input: 3
Output: 010
Explanation : Here, 010 string consists equal count of ’01’ and ’10’
subsequence and it consists at least one occurrence of 0 as well as 1.
Naive approach:
The idea is to generate every possible binary string(string which consists of only 0’s and 1’s) and find a substring with equal number of ’01’ and ’10’ subsequence.
Time Complexity: O(2N * 2N) = O(4N) because each string also has 2N subsequences.
Auxiliary Space: O(N)
Efficient approach: To solve the problem follow the below observations:
Case 1 (If N is even): Then add 1s at N/2 and (N/2 – 1) index, and add 0s to the rest of the string. So the count of ’10’ and ’01’ will be same which is equal to 2*(N/2 – 1) = (N – 2) and string also consists of at least once occurrence of 0 and 1.
For example:
- If N = 4, that means N is even, then add 1 at index N/2 =2 and (N/2)-1 = 1 .
- And add 0s at remaining positions from 0 to 3 .
- Hence, string is “0110”.
Case 2 (If N is odd): Then add 1s at middle and add 0s to the rest of the string. So the count of ’10’ and ’01’ is same which is equal to N/2 and string also consists at least once occurrence of 0 and 1.
For example:
- If N=3, that means N is odd then add 1 at index N/2=1.
- And add 0s at remaining positions from 0 to 2.
- Hence, string is “010”.
Follow the below steps to implement the above approach:
- Take an empty string s.
- Now check if N is even or odd:
- If N is even, iterate from 0 to N – 1 and add ‘1’ only at the N/2 and (N/2 – 1) index and ‘0’ on the other indices of the string.
- If N is odd, iterate from 0 to N-1 and add ‘1’ only at the N/2 index and ‘0’ in all other indices of the string.
- Now, return the constructed string.
Below is the implementation of the above approach.
C++
// C++ code to implement the approach#include <bits/stdc++.h>using namespace std;string makeString(int n){ string s = ""; // If n is even then follow this block int mid = n / 2; if (n % 2 == 0) { for (int i = 0; i < n; i++) { // If i is at middle of n and // previous middle of n // then add 1 into the string if (i == mid || i == mid - 1) { s += "1"; } // Otherwise, add 0 to the string else { s += "0"; } } } // If n is odd then follow this block else { for (int i = 0; i < n; i++) { // If i is equal to mid then // add 1 to the string if (i == mid) { s += "1"; } // Otherwise add 0 to the string else { s += "0"; } } // Return the constructed string return s; }}// Driver codeint main(){ int N = 4; // Function call cout << makeString(N); return 0;} |
Java
// Java code to implement the approachimport java.util.*;class GFG{static String makeString(int n){ String s = ""; // If n is even then follow this block int mid = n / 2; if (n % 2 == 0) { for (int i = 0; i < n; i++) { // If i is at middle of n and // previous middle of n // then add 1 into the String if (i == mid || i == mid - 1) { s += "1"; } // Otherwise, add 0 to the String else { s += "0"; } } } // If n is odd then follow this block else { for (int i = 0; i < n; i++) { // If i is equal to mid then // add 1 to the String if (i == mid) { s += "1"; } // Otherwise add 0 to the String else { s += "0"; } } // Return the constructed String return s; } return s;}// Driver codepublic static void main(String[] args){ int N = 4; // Function call System.out.print(makeString(N));}}// This code contributed by shikhasingrajput |
Python3
# Python code to implement the approachdef makeString(n): s = "" # If n is even then follow this block mid = n/2 if(n % 2 == 0): for i in range(n): # If i is at middle of n and previous # middle of n then add 1 into the string if(i == mid or i == mid-1): s += "1" # Otherwise, add 0 to the string. else: s += "0" # If n is odd then follow this block else: for i in range(n): # If i is equal to mid then add 1 to the string if i == mid: s += "1" # Otherwise add 0 to the string else: s += "0" # Return the constructed String return s return sN = 4# Function callprint(makeString(N))# This code is contributed by lokeshmvs21. |
C#
// C# code to implement the approachusing System;public class GFG{static String makeString(int n){ string s = ""; // If n is even then follow this block int mid = n / 2; if (n % 2 == 0) { for (int i = 0; i < n; i++) { // If i is at middle of n and // previous middle of n // then add 1 into the String if (i == mid || i == mid - 1) { s += "1"; } // Otherwise, add 0 to the String else { s += "0"; } } } // If n is odd then follow this block else { for (int i = 0; i < n; i++) { // If i is equal to mid then // add 1 to the String if (i == mid) { s += "1"; } // Otherwise add 0 to the String else { s += "0"; } } // Return the constructed String return s; } return s;}// Driver codepublic static void Main(string[] args){ int N = 4; // Function call Console.WriteLine(makeString(N));}}// This code contributed by AnkThon |
Javascript
<script>// Javascript program for above approachfunction makeString(n){ let s = ""; // If n is even then follow this block let mid = n / 2; if (n % 2 == 0) { for (let i = 0; i < n; i++) { // If i is at middle of n and // previous middle of n // then add 1 into the String if (i == mid || i == mid - 1) { s += "1"; } // Otherwise, add 0 to the String else { s += "0"; } } } // If n is odd then follow this block else { for (let i = 0; i < n; i++) { // If i is equal to mid then // add 1 to the String if (i == mid) { s += "1"; } // Otherwise add 0 to the String else { s += "0"; } } // Return the constructed String return s; } return s;}// Driver Code let N = 4; // Function call document.write(makeString(N)); // This code is contributed by code_hunt.</script> |
0110
Time Complexity: O(N)
Auxiliary Space: O(N)
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